Kepler s Laws. force = mass acceleration = m q. m q = GmM q

Transcription

1 Keper s Laws I. The equation of motion We consider the motion of a point mass under the infuence of a gravitationa fied created by point mass that is fixed at the origin. Newton s aws give the basic equation of motion for such a system. We denote by qt) the position of the movabe point mass at time t, by m the mass of the movabe point mass, and by M the mass of the point mass fixed at the origin. By Newton s aw on gravition, the force exerted by the fixed mass on the movabe mass is in the direction of the vector qt). It is proportiona to the product of the two masses and the inverse of the square of the distances between the two masses. The proportionaity constant is the gravitationa constant G. In formui force = GmM q q 3 Newton s second aw states that From these two equation one gets Dividing by m, force = mass acceeration = m q m q = GmM q q 3 q = q q 3 ) with = GM. This is the basic equation of motion. It is an ordinary differentia equation, so the motion is uniquey determined by initia point and initia veocity. In particuar qt) wi aways ie in the inear subspace of IR 3 spanned by the initia position and initia veocity. This inear subspace wi in genera have dimension two, and in any case has dimension at most two. II. Statement of Keper s Laws Johannes Keper ) had stated three aws about panetary motion. We state these aws beow. Isaac Newton ) showed that these aws are consequences ) ) For a discussion of the amount of rigour in Newton s Principia, see [Pourciau]. In fact, Newton seems to have been more interested in the direct probem of finding basic aws consistent with the Keper aws; see [Brackenridge].

2 of the basic equation of motion ). We first state Keper s aws in an informa way, then discuss a more rigid formuation, and then give various proofs of the fact that ) impies these aws. Keper s first aw: Let qt) be a maxima soution of the basic equation of motion. Its orbit is either an eipse which has one foca point at the the origin, a branch of a hyperboa which has one foca point at the origin, a paraboa whose foca point is the origin, or an open ray emanating from the origin. Keper s second aw Equa areas in equa times): The area swept out by the vector joining the origin to the point qt) in a given time is proportiona to the time. Keper s third aw: The squares of the periods of the panets are proportiona to the cubes of their semimajor axes. We now comment on the terms used in the formuation of Keper s aws. The Picard Lindeöf theorem about existence and uniqueness of soutions appies to the differentia equation ). It impies that for any soution qt) defined on an interva I, there is a unique maxima open interva I containing I such that the soution can be extended to I. Such a soution we ca maxima. I may be the whoe rea axis or have boundary points. If t 0 is a boundary point then im t I,t t 0 qt) = 0 or im t I,t t 0 qt) =. The orbit of the soution is by definition the set { qt) t I } in IR 2. One of the standard definitions of eipses is the foowing: Let F and F be two points in the pane. Fix a ength, say 2a, which is bigger or equa to the distance between F and F. Then the curve consisting of a points P for which the sum of the distances from P to F and from P to F is equa to 2a is caed an eipse with foca points F, F. a is caed the the semimajor axis of the eipse. Simiary, fix a ength 2a > 0, which is smaer than the distance between F and F. Then the curve consisting of a points P for which the absoute vaue of the difference of the distances from P to F and from P to F is equa to 2a is caed a hyperboa with foca points F, F. Finay, fix a point F and a ine g that does not contain F. Then the curve consisting of a points P for which the distance from P to the point F is equa to the distance from P to the ine g is caed a paraboa with foca point F. 2

3 Eipses, paraboas and hyperboas make up the conic sections. There are many other ways to describe conic sections. see the appendix beow and the references cited therein. For two vectors v = v,v 2 ) and w = w,w 2 ) in the pane, we denote, by abuse of notation, by v w = v w 2 v 2 w the third component of the cross product of v and w. Its absoute vaue is the doube of the area of the triange spanned by the points 0, v and v +w. Now et qt), t a,b) be a differentiabe curve in the pane. Using Riemann sums, one sees that the area swept out by the vector joining the origin 0 to the point qt) in the time between between t and t 2, a < t t 2 < b, is equa to t2 2 t qt) qt)dt. KeperssecondawstatesthatthereisaproportionaityconstantLsuchthat t 2 t qt) qt)dt = Lt 2 t ). By the fundamenta theorem of cacuus, this equivaent to saying that qt) qt) = L 2) is constant. Up to a constant depending on the mass of the partice, the quantity qt) qt) is the anguar momentum vector of the partice with respect to the origin. Thus, Keper s second aw is a consequence of the principe of conservation of anguar momentum. When the orbit of the soution is an eipse, we tak of panetary motion. In this case it foows from Keper s second aw that the motion is periodic. The period is the minima T > 0 such that qt+t) = qt) for a t IR. The precise form of the third aw is, that where a is the major semiaxis of the eipse. T 2 = 4π2 a 3 III. Proofs of Keper s Laws The proof of Keper s second aw is straightforward. By ) d dt qt) qt) = qt) qt)+qt) qt) = 0 qt) 3 qt) qt) = 0 so that qt) qt) is constant. As discussed in the previous section, this is the content of Keper s second aw. 3

4 Indeed, Another conserved quantity is the tota energy E = 2 q 2 q 3) d dt 2 q 2 ) q = d ) dt 2 q q q q) = q q + q q = q q + /2 q q) 3/2 q q ) = 0 3 by Keper s equation. The proof of the first aw is not as obvious as that of the second aw. We give severa proofs, aways using Keper s second aw and conservation of energy. Let L = q q be the constant of the second aw the anguar momentum). We assume from now on that L 0; otherwise one has motion on a ray emanating from the origin. Poar coordinates Keper s equation is rotation symmetric. Therefore it is a natura idea to use poar coordinates in the pane where the motion of the partice takes pace. Without oss of generaity we may assume that this is the q,q 2 ) pane. The poar coordinates r,ϕ of a point are defined by q = rcosϕ, r = q 2 = rsinϕ, tanϕ = q 2 q q 2 +q2 2 We first express the absoute vaue of the anguar momentum L and the energy in poar coordinates. Observe that q q = q q 2 q 2 q = rṙ cosϕ sinϕ+r2 ϕ cos 2 ϕ rṙ sinϕ cosϕ+r 2 ϕ sin 2 ϕ = r 2 ϕ Without oss of generaity we assume from now on that ϕ > 0. So Aso observe that L = r 2 ϕ, or, equivaenty, ϕ = L r 2 4) q 2 = d dt rcosϕ) 2 + d dt rsinϕ) 2 = ṙ 2 cos 2 ϕ+r 2 ϕ 2 sin 2 ϕ+ṙ 2 sin 2 ϕ+r 2 ϕ 2 cos 2 ϕ 5) = ṙ 2 +r 2 ϕ 2 4

5 By 5) and 4) the tota energy 3) is by 4) so that E = 2 q 2 r = 2ṙ2 +r 2 ϕ 2) r = 2 ṙ2 ) + L 2 r 2 r ṙ 2 = 2E + 2 r L 2 r 2 6) 6) is a differentia equation for r as a function of t. It is difficut to sove expicity and we do not do this here. Instead we derive a differentia equation for the ange ϕ as a function of the radius r. Observe that by 4). Inserting 6) gives with Therefore ϕ = dϕ dr = dr r 2 e 2 2 L = r 2 2E+ 2 r L 2 r 2 r e = ) 2 = dϕ dr = dϕ dt dt dr = ϕ ṙ = L r 2 ṙ L r 2 2E+ 2 L ) 2 = L 2 r L r 2 e 2 2 ) 2 r + 2E L 2 2, = L 2 7) Thus, with an integration constant ϕ 0 or er e ) 2 er dr = 2 ϕ ϕ 0 = arccos e r = r ) er e d ) 2 dr er e) ) dr +ecosϕ ϕ 0 ) 8) This is the equation of a conic section with eccentricity e. See Appendix A beow. If the energy E is negative, the eccentricity e is smaer than one and we have an eipse. Simiary, we for E = 0 we have e = and the conic section is a paraboa. Finay, if E > 0, we have e > and the conic section is a hyperboa. This ends the proof of Keper s first aw using poar coordinates. Observe that we used ony the equations 4) and 6) of conservation of anguar momentum and of energy. Using the formui from Appendix A, we see that in the case of eipses E < 0) the major axis is a = e 2 = 2 E 9) 5

6 since e 2 = 2 E L 2 2 *. The minor axis is b = a e 2 = 2 E 2 E L 2 = L = a L 2 2 E 0) Therefore the area of the eipse is πab = πa 3/2 L. Now et T be the period of the orbit. By Keper s second aw the the area swept out after time T is T qt) qt)dt = 2 2 L T 0 On the other hand this is equa to the area of the eipse. Therefore we get L T = πa3/2 L 2 or T = 2π a 3/2 This is Keper s third aw. The previous proof of Keper s first aw is based on the conservation aws for energy and anguar momentum and on the idea of writing the ange ϕ as a function of the radius r. In a variant of this proof one writes σ = as a function of ϕ. By the change of variabes r formua in differentia cacuus and 5) q 2 = ṙ 2 +r 2 ϕ 2 = ϕ 2 ṙ ϕ )2 +r 2) = L 2 r dr 4 dϕ )2 +r 2) = L 2 ) dr r 2 dϕ )2 + r 2 = L 2 ) d dϕ r )2 + r 2 Then by 3) E = 2 q 2 q = L 2 2 dσ dϕ )2 +σ 2) σ Differentiating with respect to ϕ and dividing by L 2 gives 0 = dσ d 2 σ dϕ dϕ +σ ) 2 L 2 Since dσ dϕ 0 this impies d 2 σ +σ = 0 dϕ 2 L 2 * It is remarkabe that a depends ony on the energy. So a bounded orbits with the same energy are eipses with the same ength of the major axis, but of course their position in space is ddifferent because the eccentricity then varies with L 6

7 The genera soution of this differentia equation is σ = L 2 +e cosϕ ϕ0 ) ) with integration constants e, ϕ 0. Remembering that r = σ we get again the equation r = for the orbit. +ecosϕ ϕ 0 ) and setting = L 2 as in 7) In this approach just comes as an integration constant, but now it can easiy be identified with the quantity of 7). The Lapace Lenz Runge vector The basic equation of motion is a second order differentia equation. Its soution is competey detemined by the position q and veocity q at any given time t. Therefore one expects that the quantities characterizing the orbits can be expressed in terms of q and q. For the eccentricity of the conic section this is done in 7), since E and L are expressed in terms of q and q in 3) and 2). Simiary, in the case of eipses, the ength of the major and minor axis are described in terms of q and q by 9) and 0). If one considers the Keper probem as a probem in three dimension, the pane of the orbit is determined as the pane through the origin perpendicuar to the anguar momentum vector L. These data determine the Keper eipse up to rotation around the origin the integration constant ϕ 0 of the previous subsection. To fix this ambiguity we woud ike to find a vector in the direction of the major axis that can be expressed purey in terms of q and q. The standard choice is the Lapace Lenz Runge vector A = q q + q L = q 2 ) q q q q) q ) Here we used the identity x y z) = x y)z +x z)y for vectors x,y,z IR 3 to see that q L = q q q) = q q) q + q 2 q. Observe that, for a vector v in the invariant pane orthogona to L, v L is the vector in this pane obtained from v by L rotating by 90 o. So L q L is the vector obtained from q by rotating by 90o. First we verify that A reay is constant during a Keper motion. Using the basic equation of motion ) and the definition of L da dt = q q + q q q 3 q + q L = q q + q q q 3 q q 3 q q q) = 0 We aso used the fact that d dt q = d dt q q) /2 = q q)q q) 3/2 and again the identity x y z) = x y)z +x z)y. 7

8 Next we caim that the ength of the Lapace Lenz Runge vector is equa to the eccentricity e. For this reason A is caed the eccentricity vector in [Cushman]. To prove the statement about the ength of A observe that A A = 2 q q q L)+ 2 q L 2 Since q and L = q q are perpendicuar, q L 2 = q 2 L 2. By the standard vector identity x y z) = z x y) Therefore by 3). q q L) = L q q) = L L = L 2 2) A A = 2 2 q L q 2 L 2 = + L 2 2 q q = +2 L 2 2 E = e 2 ) Before using the Lapace Lenz Runge vector, we describe how one coud get the idea to consider it, once one aready knows Keper s aws*. Let us consider the case of negative energy, so that the orbits are eipses. A natura invariant of the eipse is the vector joining the two foci. The origin is one focus of the eipse, ca the other one f. It is known that for each point q of the eipse, the ines joining q to the origin and q to the other focus f form opposite equa anges with the tangent ine of the eipse. See Appendix A. Therefore a vector from q in the direction of f is obtained by adding to q twice the orthogona projection of q to the tangent direction of the eipse at q. q A unit tangent vector to the eipse at the point q is q. Therefore f q has the direction q +2 ) q q q. This means that there is a scaar function αt) such that q q ) f = qt)+αt) qt) 3) qt)+ 2qt) qt) qt) 2 for a t. Differentiating and using Keper s aw ) gives 0 = f = q + α q +2 q q q q ) [ +α q +2 q q+q q) 2 q 2 q q)2 q q) q 4 ] ) q + 2q q q q 2 = αq + q 2 { q 2 +2 αq q)+α [ q 2 +2 q 2 +2q q q 3 ) 2 q q q 2 q q q 3 ) ]} q = [ α+2α +2α q q q 2 q q 3 ) ] q q q 2 q q + { q 2 +2q q) [ α+2α ] [ q q 3 q 2 q 2 q +2α 3 2 q 2 q ] } q * For other arguments, see [Heintz] and [Kapan]. Remarks on the history of this vector can be found in [Godstein] and [Godstein]. 8

9 Since q and q are ineary independent, this impies that the coefficients of q and q are both zero. From the coefficient of q we get α+2α q q q 2 q 3 = 0 Inserting this, and the fact that 2 q 2 q = E into the coefficient of q gives So α = q 2 2E q 2 +2αE = 0. Inserting this into 3) gives ) f = q q 2 2E q + 2q q q q 2 E + q ) ) 2 q q q) q = E = E = E A 2 q 2 ) ) q q q q) q 4) The argument started with the observation that f shoud be a conserved quantity. As E is conserved, this suggests that A is a conserved quantity. Above, we have proven this directy. Observe from 9), that in the case of eipses, is twice the major semiaxis a of the eipse. This is consistent with the fact that the distance between the two foci is 2ea. E Once one knows that the Lapace Lenz Runge vector and the anguar momentum vector are constants of the Keper motion, the proof of Keper s second aw is reativey fast. As q A = q q q + q L) = q + q q L) = q + L q q) = q + L 2 we have, setting again e = A q = e ) e L 2 A q A The expression in brackets is the distance from q to the ine perpendicuar to A through the point L 2 e 2 A. If we ca this ine the directrix, then the equation above states that the ratio of the distance between q and the origin and the distance between q and the directrix is equa to e. As pointed out in Appendix A, this is one of the characterizing properties of conic sections. Exercise: In the case of negative energy, show that q f + q is constant! Here, f is the vector of 4). 9

10 Taking the cross product of the Lapace Lenz Runge vector A with the anguar momentum vector L gives L A = q L q)+ L q L) = q q L+ L 2 q 5) using the vector identity x y z) = x y)z+x z)y and the fact that q and L are perpendicuar. Therefore q = L 2 L A q q L) = L 2 L A+ L 2 L q q This equation determines the veocity vector q of the Keper motion as a function of the position q. Observe that L L A is independent of t and that is aways a vector 2 of ength one perpendicuar to L. So L q L 2 q is aways a vector of ength L in the pane perpendicuar to L. Consequenty, for a fixed Keper orbit, the veocity vectors a ie on the circe around the point momentum hodograph. A momentum space argument L 2 L A of radius q q. This circe is caed the L In this subsection, we prove Keper s first aw, starting with the anaysis of the hodograph*, that is the curve traced out by the momentum vector pt) = qt) The equation of motion ) is so that Consequenty ṗ = q q 3 p = 3q q q + q 3 q q 5 ṗ p = 2 q 6 q q = 2 q 6 L The standard formua for the curvature of a pane curve shows that the curvature of the hodograph at the point q is κ = ṗ p ṗ = 2 L q 6 3 q 6 3 = L * Another derivation of the equation of the hodograph, attributed to Hamiton, can be found in [Hankins], ch.24. 0

11 This proves that the curvature of the hodograph is constant. So it is a circe. Its radius is κ = L. At each of its points p, the vector pointing to the center u of the circe is perpendicuar to L and to the tangent vector ṗ. Its ength is the radius L. So and our argument shows that p u = L u = p+ is a conserved quantity ). The hodograph has the equation ṗ ṗ L L = q L 2 q L 6) q L 2 q L 7) p u = L So the hodograph is the circe around u of radius e = + 2E L 2 as in 7), since 2 e. Observe that u = L L where 2 u 2 = p 2 + q L 2 +2 p q L) q 2 L 4 q L 2 = p L 2 +2 = p L 2 +2 q L 2 L p q) q ) 2 = 2E + 2 L 2 = e L We now describe the position q in terms of the momentum p. By 6) and the fact that q is perpendicuar to to q L, we have q p u) = 0. So q is perpendicuar to p-u) and therefore it is a mutipe of p u) L. Write q = r p u) L p u) L = r p u) L with r = ± q As pointed out in 2), q p L) = L 2. If we insert the representation of q above, we get r p u) L ) p L) = L 2 Cross product by L L corresponds to rotation by 90o. Therefore this impies = rp u) p = r p u) p u)+p u) u ) = r 2 L e L 2 p u) p u u u ) From 5) one sees that u = L 2 L A, where A is the Lapace Lenz Runge vector of the previous subsection. Conversey, A = u L. One can prove directy that u is a conserved quantity and then derive the Lapace Lenz Runge vector using this formua. )

12 Here we used that p u = L p u and u by ϕ, we get e and that u = L. If we denote the ange between r = +e cosϕ with = L 2 as in 7). So we obtain again the equation of the conic as in Appendix A. A purey geometric proof of Keper s first aw using the hodograph was given by Hamiton, Kevin and Tait, Maxwe, Fano and Feynman independenty; see Feynman s ost ecture [Goodstein]. It first describes the hodograph in geometric terms and then deduces the Keper orbit as the enveoppe of its tangent ines. This argument is in parts very cose to Newton s origina argument. There is a debate whether Newton s origina argument meets the 2 st century standards of rigour. For references see the introduction of [Derbes]. [Brackenridge] provides a guide and historica perspective on Newton s treatment of the Keper probem. The eccentric anomay In the proofs of Keper s first aw given above we derived the shape of the orbit, but did not actuay write down a parametrization by the time t. Such a parametrization woud correspond to a soution of the basic equation of motion ). However, 8) is a parametrization of the orbit in terms of the ange ϕ ϕ 0 aso caed the true anomay, see Appendix A), and conservation of anguar momentum 4) impicity gives the dependence of ϕ on t by the differentia equation dϕ dt = L. To simpify the discussion we put ϕ r 2 0 = 0. First we consider the case of negative energy, that is, the case of bounded orbits. Write E = ε2 2 In this case the orbit is described by the equation q +ea) 2 a 2 + q2 2 b 2 = where the principa axes a,b of the eipse and the eccentricity e are determined by 9), 0) and 7) respectivey. We use the parametrization q = a coss ea, q 2 = b sins by the eccentric anomay s see Appendix A). To get the dependence of s on the time t, observe that by formua 24) of the appendix and 7) ds dϕ = e 2 r = 2E L r = ε L r 2

13 We just observed that dϕ dt = L r 2. Therefore, by the chain rue ds dt = ε r 8) By9), ε 2 = 2 E = a Hence 8) gives or, equivaenty. Informua 25)ofthe appendix, we show that r = a ecoss). ds dt = a 3/2 ecoss) dt ds = a3/2 ecoss) Integrating both sides with respect to s gives Keper s equation t t 0 = a3/2 s e sins) 9) Keper s equation is an impicit equation for s as a function of t. It is not a differentia equation anymore, but just an equation that invoves the inversion of the eementary function s s sins. This inversion cannot be performed by eementary functions. Oneoftheadvantagesoftheeccentricanomayis, thatitiswesuitedforadescription of the Keper motion in position space. Therefore we derive the equation of motion with respect to this parameter. We make the change of variabes 8) in the basic equation of motion ). Reca that r = q. Then Therefore, using ) dq ds dt = q ds = q ε dq q or, equivaenty q = ε q ds d 2 q ds = d q q d q 2 ε ds q + ε ds = d q dq q ds ds + q 2 ε q = d q dq 2 q ds ds ε 2 q q = q q dq 2 ds ) dq ds ε 2 q q 20) Once one knows that 8) is a good change of variabes for the Keper probem and that the Lapace Lenz Runge vector is a conserved quantity, one can give a quick proof of Keper s first aw: By 3) and ), the Lapace Lenz Runge vector is ) A = 2E + q q q q) q = ε2 q q dq 2 ds = ε2 q q dq 2 ds ) dq ds 2E ε 2 q ε 2 q q ) ) dq ds ε 2 q q +q ) 3

14 since ε 2 = 2E. Therefore 20) gives d 2 q ds 2 +q = ε 2 A The genera soution of this second order inhomogenuous inear differentia equation with constant coefficients is qs) = C coss+c 2 sins ε 2 A 2) with constant vectors C,C 2. This is the parametrization of an eipse. The vectors C,C 2 are easiy described in terms of known quantities, see [Cordani]. In the parametrization 2) a orbits have period 2π. This is even true for the orbits with zero anguar momentum in this case C and C 2 are ineary dependent), in contrast to the true Keper fow where the point mass crashes into the origin. One says that the eccentric anomay reguarizes the coisions. Appendix A: Conic sections Conic sections are the nonsinguar curves that are obtained by intersecting a quadratic cone with a pane. The reation with the foca description given after the statement of Keper s aws can be seen using the Dandein spheres, see [Knörrer] Aso, conic sections are the zero sets of quadratic poynomias in two variabes that do not contain a ine. We discuss the most reevant properties of the conic sections Eipses Consider the eipse consisting of a points P for which the sum of the distances P F and P F is equa to 2a. The eccentricity e of the eipse is defined by F F = 2ea Ceary, 0 e <. Let M be the midpoint between the two foci the center of the eipse). The ine through the two foci is caed the major axis of the eipse. The two points of the eipse on the major axis have each distance a from the center. The foci both have distance ea from the center. The ine through M perpendicuar to the major axis is caed the minor axis. It is the perpendicuar bisector of the two foci. The points of the eipse on the minor axis have distance a from both foci. It foows from Pythagoras theorem that the distance of these points from the center is b = a e 2 4

15 If one introduces Cartesian coordinates x,y centered at M with the major axis as x axis and the minor axis as y axis then the equation of the eipse is x 2 a 2 + y2 b 2 = 22) For a proof see [Knörrer], Satz 4.3. The area encosed by the eipoid is equa to πab; see for exampe [Zorich], 6.4.3, Exampe 5. An important consequence of the foca description of the eipse is the foowing: For each point q of the eipse, the ines joining q to the origin and q to the other focus f form opposite equa anges with the tangent ine of the eipse. For a proof, see [Knörrer], Satz 4.6. Another usefu description of the eipse is the foowing. The ine perpendicuar to the major axis that has distance a to the center and ies on the same side of the center e as the focus F is caed the directrix with respect to F. One can show that the eipse is the set of points P for which the ratio of the distance from P to F to the distance from P to the directrix is equa to the eccentricity e. To see this, note that a point x,y) fufis the directrix condition with focus ea,0) if and ony if x ea) 2 +y 2 = e 2 x a e )2 x 2 a 2 + y2 e 2 )a = 2 This description directy gives the equation of the eipse in poar coordinates ) r,ϕ) centered at the focus F. Choose the anguar variabe ϕ such that ϕ = 0 corresponds to the ray starting at F in the direction away from the center ). If a point P has poar coordinates r,ϕ) then its distance to F is r. Since the distance from F to the directrix is e e) a, the distance from P to the directrix is e e) a rcosϕ. Thus the equation of the eipse is ) r = e e e) a rcosϕ or, equivaenty r = +e cos ϕ 23) where = e 2 )a. The quantity is caed the parameter of the eipse. The ange ϕ is caed the true anomay of a point on the eipse with respect to the focus F. Observe that ϕ = 0 corresponds to the point of the eipse cosest to F. In ceestia mechanics, this point is caed the periheion. Formua 23) is a parametrization of the eipse, giving the distance r to the focus as a function of the true anomay ϕ. Equation 22) suggests the parametrization x = a coss, y = b sins ) For another derivation, see [Knörrer] 4.7. ) This is the direction from the focus F to the cosest point on the eipse. 5

16 of the eipse. The parameter s is caed the eccentric anomay. To get the reation between the eccentric amnomay and the true anomay with respect to the focus F = ea,0), et x,y) be a point of the eipse with true anomay ϕ and eccentric anomay s. Then x = a coss y = b sins = a e 2 sins and x = rcosϕ +ea = y = rsinϕ = e2 )a +ecosϕ sinϕ +ecosϕ cosϕ +ea = e2 )a +ecosϕ cosϕ +ea Comparing these two representation, we obtain coss = e+ e2 +ecosϕ cosϕ, sins = e 2 +ecosϕ sinϕ We differentiate the first equation with respect to ϕ and insert the second to get ds dϕ sins = e2 +ecosϕ sinϕ+ e 2 +ecosϕ) 2 esinϕ cosϕ = e 2 sins+ e 2 ecosϕ +ecosϕ sins = e 2 sins Dividing by sins and using 23) gives +ecosϕ We aso need the expression of r in terms of s. Since ds dϕ = e 2 r 24) r 2 = x ea) 2 +y 2 = a 2 coss e) 2 + e 2 )a 2 sin 2 s = a 2 cos 2 s 2ecoss+e 2 + e 2 ) cos 2 s+e 2 cos 2 s ) = a 2 ecoss) 2 we have r = a ecoss) 25) Hyperboi Consider now the hyperboa consisting of a points P for which the difference of the distances P F and P F has absoute vaue equa to 2a. The eccentricity e of the hyperboa is defined by F F = 2ea 6

17 Ceary, e >. As before, et M be the midpoint between the two foci caed the center). The ine through the two foci is caed the major axis of the hyperboa. The two points of the hyperboa on the major axis have each distance a from the center. The foci both have distance ea from the center. If one introduces Cartesian coordinates x,y centered at M with the major axis as x axis then the equation of the eipse is where b = a e 2. x 2 a 2 y2 b 2 = The description using a directrix is amost identica to the one for eipses. The ine perpendicuar to the major axis that has distance e a to the center and ies on the same side of the center as the focus F is caed the directrix with respect to F. The hyperboa is the set of points P for which the ratio of the distance from P to F to the distance from P to the directrix is equa to the eccentricity e. In poar coordinates centered at F one gets as equation for the hyperboa r = e2 )a +e cos ϕ 26) Here, the anguar variabe ϕ has been chosen such that ϕ = 0 corresponds to the ray starting at F in the direction to the center ). Paraboi Finay consider the paraboa consisting of a points that have equa distance form the focus F and the ine g which we ca the directrix of the paraboa Let be the distance from F to g. Furthermore et be the ine perpendicuar to g through F and M the midpoint of its segment between F and g. It has distance /2 both from F and g. FIGURE If one introduces Cartesian coordinates x, y centered at M with the x axis being oriented in direction of M then the equation of the paraboa is y 2 +2x = 0 Again we choose poar coordinates centered at F such that ϕ = 0 corresponds to the ray from F to M. As before one sees that the equation of the paraboa is r = +cos ϕ 27) ) Again, this is the direction from the focus F to the cosest point on the eipse. 7

18 Appendix B: The two body probem The basic equation of motion ) aso governs the genera two body probem. Here we consider two point masses with masses m,m 2 and time dependent positions r t),r 2 t). In this situation, Newton s aws give FIGURE r m r t) = Gm m 2 t) r t) 2 r 2 t) r t) 3 28) r m 2 r 2 t) = Gm m t) r 2 t) 2 r t) r 2 t) 3 Denote by Rt) = m +m m 2 r t) + m 2 r 2 t) ) the center of gravity. Adding the two equations of 28) gives Rt) = 0. That is, the center of gravity moves with constant speed. Aso, set qt) = r 2 t) r t). 28) gives r t) = Gm 2 qt) qt) 3, r 2 t) = Gm qt) qt) 3 so that This is the basic equation ). qt) = qt) qt) 3 Appendix C: The Lagrangian formaism The Euer Lagrange equations associated to a function Lt, q, p) is d L ) dt p i p= q L q i p= q = 0 29) Conventionay one views L as a function of the variabes t,q, q and uses the shorthand formuation d L ) dt q i L q i = 0 The Lagrange function for the Keper probem is the difference between the cinetic and the potentia energy It is independent of t. Then L K q,p) = 2 p 2 + q L K p i = p i, d LK ) dt p i p= q = qi, 8 L K q i = q i q 3

19 so that the associated Euer Lagrange equation are q i + q i q 3 = 0 This is the basic equation of motion ). The Euer Lagrange are reated to the foowing variationa probem. Fix points q,q IR 3 and times t < t. For each twice differentiabe curve define the action γ : [t,t ] IR 3 t qt) with qt ) = p, qt ) = q Φγ) = t t L t,qt), qt) ) dt If γ minimizes the action Φγ) among a curves as above, then the Euer Lagrange equations hod. See [Arnod, Section 3]. The fact that the equations of motion of an autonomous mechanica system are minimizers of the action cinetic energy) - potentia energy) is caed the principe of east action or principe of Maupertuis). Another instance of a variationa probem is the construction of geodesics. Let U be an open subset of IR n. Assume that for each point q U one is given a positive definite symmetric biinear form on IR n a Riemannian metric). It is represented by a positive symmetric n n matrix Gq) = g ab q) ). For simpicity we assume that the a,b=,,n coefficients g ab q) are C functions of q. The ength of a curve γ : [t,t ] U, t qt) with respect to the Riemannian metric is by definition engthγ) = t t qt) G qt) ) ) 2 qt) dt By definition geodesics are ocay shortest connections. That is, a curve t qt) is a geodesic if and ony if, for each t, there is ε > 0 such that for a s with t < s < t+ε, the curve is a shortest connection between the points qt) and qs). It foows that geodesics fufi the Euer Lagrange equations 29) with Lq,p) = n p Gq)p = a,b= g ab q)p a p b ) 2 Obviousy the ength of a curve does not change under reparametrization. In particuar, the critica point for the variationa probem is degenerate. To normaize the situation, we ook for minimizing curves that are parametrized with constant speed different from zero). That is, curves which are parametrized in such a way that qt) G qt) ) qt) = const for 9

20 a t. If one has a minimizer for the variationa probem that is parametrized by arcength, then it aso fufis the equation d L 2 ) dt p i p= q L 2 q i p= q = 0 30) Indeed, mutipying the Euer Lagrange equation 29) by const = Lq, q), we get [ ) ] d const Lq, q) L dt p i Lq, q) L q p= q i p= q = 0 which impies 30). Equation 30) is simper than the Euer Lagrange equations associated to the origina Lagrange function. We now derive the equation of motion associated to 30). To see that we reay get geodesics, we then have to verify that the resuting curves are indeed parametrized with constant speed. Equation 30) gives 2 d n ) g dt ab q) q b n g bc q a q b q c = 0 for a =,,n b= b,c= Since d dt g abq) = n g ab c= q c q c, 30) is equivaent to n n g ab q b = g bc 2 q a q b q c n b= = 2 b,c= n b,c= b,c= g ab q c q b q c gbc q a g ab q c g ca q b ) ) q b q c for a =,,n. Here we used that g ab q c = g ba q c and exchanged the summation indices b and c. The equations above state that the a component of the vector G q is equa to the n gbc q a g ab q c g ca q b ) ) q b q c. We denote by g ab the entries vector with entries 2 b,c= of the inverse matrix G. Then the equations are equivaent to where Γ a bc q a + n b,c= are the Christoffe symbos n Γ a bc = n 2 g ad d= b,c= Γ a bc q b q c = 0 3) gdb q c + g cd q b g ) bc q d To verify that 3) reay describes geodesics, we sti have to verify that soutions of 3) are curves that are parametrized by constant speed. So et t qt) be a soution of 3). Reversing the cacuation above, we see that shows 2G q is the vector with entries n gbc b,c= q a g ab q c g ) ca q b qb q c. Therefore d qgq) q = 2 qgq) q + qġ q dt = n a,b,c= and qgq) q is indeed constant. gbc q a g ab q c g ca q b ) qa q b q c + n 20 a,b,c= g bc q a q a q b q c = 0